Discussion in 'Math' started by Khalid Abur-Rahman, Feb 11, 2009.

1. ### Khalid Abur-Rahman Thread Starter Active Member

Dec 25, 2008
34
0
Can someone show me how to put this equation into quadratic forum?

its specifaclly for JFET design by the way (Common Source Amplifier)

A(v) = -(20k||R(d)/3k-R(d)+350=-4

where k is for kilo ohms..

how do I put it in quadratic form?

Can you show me step by step?

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
I am not sure where the -4 comes in, is this the numerical value of the amplifcation?

$\frac{20000*Rd}{20000+Rd}$ * $\frac{-1}{3000-Rd+350}$

If you multiply out the fractions you will get your quadratic in Rd. You should be able to carry on.

Perhaps if you set this equal to -4 you will be able solve for Rd.

3. ### nene biggie New Member

Feb 28, 2009
3
0
simplify to the simplest form then amply the quadratic formular ie x=-b+/-√b-4ac all over 2a